The thermal conductivity of a material is a measure of its ability to conduct heat. It is defined by the formula:
Q = k × A × ΔT/Δx × t
Where:
- Q is the heat energy conducted (in Joules)
- k is the thermal conductivity (in W/mK)
- A is the area through which heat is conducted (in square meters, m2)
- ΔT is the temperature difference (in Kelvin or Celsius, since a difference in degrees Celsius is equivalent to a difference in Kelvin)
- Δx is the thickness of the material (or the distance over which the temperature difference occurs, in meters)
- t is the time taken (in seconds)
We are given:
- Q = 288,000 J (since energy is in KJ and 1KJ = 1,000J)
- ΔT/Δx = 90 ºC/m (temperature gradient)
- t = 7,200 s
The cube has each side measuring 3 meters, so the area A of one face (since heat is conducted across two opposite faces, effectively using one face area for calculation) is:
A = 3m × 3m = 9 m2
Now, we need to solve for k (thermal conductivity):
Q = k × A × ΔT/Δx × t
288,000 J = k × 9 m2 × 90 ºC/m × 7,200 s
k = 288,000 / (9 × 90 × 7,200)
Calculate the denominator:
9 × 90 × 7,200 = 5,832,000
Therefore:
k = 288,000 / 5,832,000 ≈ 0.0493 W/mK
This converts approximately to 4.93 × 10-2 W/mK.
Therefore, the correct answer is 4.9 × 10-2 W/mK.